$\frac{m^3+n^3}{m^2+n^2+m+n}=800$

Find all ordered pairs of positive integers $(m, n)$ satisfying the equation above, and enter your answer as $\sum (m+n)$ .

The answer is 3200.

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We can further proceed by determining factors of N ,so as to reduce the range until we can check manually.Quite similar to what you have done with k.

Shivansh Kaul
- 3 years, 11 months ago

Did the same way as you Mark! This is a nice technique.Just narrow down the possible values by implementing the condition that root of discriminant should be less than N.Since smaller root should also be positive.

Shivansh Kaul
- 3 years, 11 months ago

Mark, exactly how did you find the value of N ? By trial and error ?

geni percaso
- 3 years, 11 months ago

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I typed

Select[Table[j,{j,534,1602}],IntegerQ[Sqrt[#(3#-1600)(3200 + 1600# - #^2)]] &]

into Mathematica. Although some Number Theory could cut the problem down to computer-independent size, checking a finite problem is something that computers are excellent at, and it saved me a huge amount of time

Mark Hennings
- 3 years, 11 months ago

I think N cannot be 1602, because then the expression 3200 + 1600N - N^2 is negative.

Scrub Lord
- 3 years, 9 months ago

How did you know to use $m<1000$ and $n<1000$ ?

Kazem Sepehrinia
- 3 years, 6 months ago

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I try it for $(m,n)=(1000,1000)$ and get total answer $>800$ .Taking higher value than $1000$ gradually the answer increase. So, I think that, there are no any value for $(m,n)$ greater than $1000$ . Actually, it's not mandatory to take $m<1000,n<1000$ . To continue the loop I used it.

Md Mehedi Hasan
- 3 years, 6 months ago

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Relevant wiki: General Diophantine Equations - Problem SolvingEquation is $m^3+n^3=800 \big(m^2+n^2+m+n \big). \qquad (1)$ First note that $m^3+n^3 \stackrel{5}{\equiv} 0$ . We claim that $m+n \stackrel{5}{\equiv} 0$ . If both of $m$ , $n$ are divisible by $5$ this is true. Let $5 \nmid m$ and $5 \nmid n$ . Then we must have either $m^3 \stackrel{5}{\equiv} 1, n^3 \stackrel{5}{\equiv} 4$ or $m^3 \stackrel{5}{\equiv} 2, n^3 \stackrel{5}{\equiv} 3$ , which in turn give $m \stackrel{5}{\equiv} 1, n \stackrel{5}{\equiv} 4$ or $m \stackrel{5}{\equiv} 3, n \stackrel{5}{\equiv} 2$ respectively. Both of cases lead to $m+n \stackrel{5}{\equiv} 0$ .

On the other hand $m^3+n^3 \equiv 0 \pmod{32}$ . Observe that $m, n$ have same parity. If $m, n$ are both odd, then $m^2-mn+n^2$ is odd and this implies that $m+n \equiv 0 \pmod{32}$ . If $m, n$ are both even, then take $m=2m_1$ and $n=2n_1$ and rewrite the equation $(1)$ as $m_1^3+n_1^3=200(2m_1^2+2n_1^2+m_1+n_1) \qquad(2)$ Therefore, $m_1, n_1$ have same parity. If $m_1, n_1$ are both odd, then $m_1+n_1 \equiv 0 \pmod{16}$ and $m+n \equiv 0 \pmod{32}$ . If $m_1, n_1$ are both even, then let $m_1=2m_2$ and $n_2=2n_2$ and rewrite the equation $(2)$ as $m_2^3+n_2^3=50(4m_2^2+4n_2^2+m_2+n_2) \qquad(3)$ If $m_2, n_2$ are both odd, then $m_2+n_2 \equiv 0 \pmod{4}$ and $m+n \equiv 0 \pmod{16}$ . If $m_1, n_1$ are both even, then let $m_2=2m_3$ and $n_2=2n_3$ and rewrite the equation $(3)$ as $2(m_3^3+n_3^3)=25(8m_3^2+8n_3^2+m_3+n_3)$ Hence $m_3+n_3 \equiv 0 \pmod{2}$ and $m+n \equiv 0 \pmod{16}$ . Therefore we have $m+n \equiv 0 \pmod{16}$ at least.

Using the conclusions $m+n \stackrel{5}{\equiv} 0$ and $m+n \stackrel{16}{\equiv} 0$ , one can write $m+n=80k$ . Equation $(1)$ gives $80k \big(6400k^2-3m(80k-m) \big)=800 \big(6400k^2-2m(80k-m)+80k \big),$ or a quadratic in terms of $m$ : $(3k-20)m^2-80k(3k-20)m+800k(8k^2-80k-1)=0,$ The discriminant is $1600k^2(3k-20)^2-800k(3k-20)(8k^2-80k-1)=800 k(3k-20) \big( -2k^2+40k+1 \big),$ which must be a square. Thus $k(3k-20)\big(-2k^2+40k+1 \big)=2l^2 \qquad (4)$ Look mod $4$ , then $2k^4+3k^2 \stackrel{4}{\equiv} 2l^2$ , which implies that $k \equiv 0 \pmod{4}$ and $l$ is even. Write $k=4K$ and $l=2L$ . Equation $(4)$ becomes $2K(3K-5)\big(-32K^2+160K+1 \big)=L^2 \qquad (5)$ Look mod $5$ , then $3K^4+K^2 \stackrel{5}{\equiv} 2L^2$ , which implies that $K \equiv 0 \pmod{5}$ or $K \equiv 2 \pmod{5}$ . From equation $(5)$ one can get $1<K<6$ , otherwise its LHS will be negative. Thus only possibilities for $K$ are $2, 5$ and its easy too see that only $K=5$ makes $L$ an integer.

Finally $m+n=80k=320K=1600$ . Solving above quadratic in terms of $m$ gives $m=820, 780$ . Therefore there are two solutions $(m, n)=(820, 780), (780, 820),$ and gives the answer as $\boxed{3200}$ .